For a 0.1 M Ba(NO_3)_2 solution, the van't | vedclass

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(C) The dissociation of $Ba(NO_3)_2$ is given by: $Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-$. Here, the number of ions produced per formula unit, $n =

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$0.87$

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(C) The dissociation of $Ba(NO_3)_2$ is given by: $Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^-$. Here, the number of ions produced per formula unit, $n = 3$. The formula for the van't Hoff factor $(i)$ in terms of the degree of dissociation $(\alpha)$ is: $i = 1 + (n - 1)\alpha$. Substituting the given values: $2.74 = 1 + (3 - 1)\alpha$ $2.74 = 1 + 2\alpha$ $1.74 = 2\alpha$ $\alpha = \frac{1.74}{2} = 0.87$. Thus, the degree of dissociation is $0.87$.

For a $0.1 \ M \ Ba(NO_3)_2$ solution, the van't Hoff factor is $2.74$. The degree of dissociation is .............

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